A block of mass 20 kg is acted upon by a force `F=30N` at an angle `53^@` with horizontal in downward direction as shown. The coefficient of friction between the block and the forizontal surface is `0.2`. The friction force acting on the block by the ground is `(g=10(m)/(s^2)`)

Options

A. `40.0N`
B. `30.0N`
C. `18.0N`
D. `44.8N`

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and calculate the frictional force.

### Step 1: Identify the given values
- Mass of the block, \( m = 20 \, \text{kg} \)
- Applied force, \( F = 30 \, \text{N} \)
- Angle of applied force, \( \theta = 53^\circ \)
- Coefficient of friction, \( \mu = 0.2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the weight of the block
The weight \( W \) of the block can be calculated using the formula:
\[ W = m \cdot g = 20 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 3: Resolve the applied force into components
The applied force \( F \) has two components:
- Horizontal component:
\[ F_x = F \cdot \cos(\theta) = 30 \, \text{N} \cdot \cos(53^\circ) \]
Using \( \cos(53^\circ) = \frac{3}{5} \):
\[ F_x = 30 \cdot \frac{3}{5} = 18 \, \text{N} \]

- Vertical component:
\[ F_y = F \cdot \sin(\theta) = 30 \, \text{N} \cdot \sin(53^\circ) \]
Using \( \sin(53^\circ) = \frac{4}{5} \):
\[ F_y = 30 \cdot \frac{4}{5} = 24 \, \text{N} \]

### Step 4: Calculate the normal force
The normal force \( N \) can be calculated by considering the vertical forces acting on the block. The normal force is equal to the weight of the block plus the vertical component of the applied force:
\[ N = W + F_y = 200 \, \text{N} + 24 \, \text{N} = 224 \, \text{N} \]

### Step 5: Calculate the maximum frictional force
The maximum frictional force \( f_{\text{max}} \) can be calculated using the formula:
\[ f_{\text{max}} = \mu \cdot N = 0.2 \cdot 224 \, \text{N} = 44.8 \, \text{N} \]

### Step 6: Determine the actual frictional force
The actual frictional force \( f \) acting on the block will be equal to the horizontal component of the applied force, as long as it does not exceed the maximum frictional force:
\[ f = F_x = 18 \, \text{N} \]

Since \( f_{\text{max}} (44.8 \, \text{N}) > F_x (18 \, \text{N}) \), the actual frictional force is:
\[ f = 18 \, \text{N} \]

### Final Answer
The friction force acting on the block by the ground is \( 18 \, \text{N} \).